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两个长整数字符串求和(不允许使用ES6+), 面试手撸代码遇到到这个问题

1. 实现方式第一种

// 短整数字符串前边补 '0'; num需要补 '0' 的短整数字符串, len 长整数字符串的长度  
function fillZero (num, len) {let str = num.toString();if (str.length < len) {str = '0'.repeat(len - str.length) + str;}return str;}// 处理相加大于等于10进制问题; num1, num2 两个长度一样的长整数字符串function handlerCarry (num1, num2) {let arr = []for (let i = 0; i < num1.length; i++) {arr.push(Number(num1[i]) + Number(num2[i]))}arr = arr.reverse()arr.forEach((item, index) => {// console.log(item)if (item >= 10) {arr[index] = item - 10arr[index + 1] = arr[index + 1] ? arr[index + 1] + 1 : 1}})// console.log(arr)return arr.reverse().join('')}// 处理两个长整数字符串相加后返回结果字符串; num1, num2 长度不一定相等的两个长整数字符串function longIntStrSum (num1, num2) {let str = ''let len1 = num1.length;let len2 = num2.length;if (len1 < len2) {num1 = fillZero(num1, len2)} else if (len1 > len2) {num2 = fillZero(num2, len1)}str = handlerCarry(num1, num2)return str}let num1 = '123412412'let num2 = '967354215'console.log(longIntStrSum(num1, num2))

2. 借鉴他人的代码实现两个长整数字符串相加

function sumStrings (a, b) {var result = [], count = 0;if (a.length < b.length) b = [a, a = b][0];b = Array(a.length - b.length + 1).join('0') + b;var arrA = a.split('');var arrB = b.split('');for (var j = 0; j < a.length; j++) {var temp = (Number(arrA.pop()) + Number(arrB.pop())) + count;temp >= 10 ? [temp, count] = [temp - 10, 1] : count = 0;result.push(temp);}result.push(count);return result.reverse().join('').replace(/^0+/, '');
}
let sum = sumStrings('1234', '456789')
console.log(sum) // 458023